Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The combustion of 
will be,
The intermediate balanced chemical reaction will be,
(1) 
(2) 
Now we are reversing reaction 1 and then adding both the equations, we get :
(1) 
(2)
The expression for enthalpy change for the reaction will be,
Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol
The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
Answer: 8.12 g NaCl
Explanation: Use Avogadro's number to find the number of m
moles of NaCl:
8.24x10²² molecules NaCl / 1 mole NaCl/ 6.022x10²³ molecules NaCl
= 0.14 mole NaCl
Next convert moles to grams NaCl using its molar mass;
0.14 mole NaCl x 58g NaCl / 1 mole NaCl
= 8.12 g NaCl
Answer:
A
Explanation:
Hydrocarbons with short chain lengths are more volatile than those with longer chains. A practical example of this can be seen in the first few members of the alkane series. They are mostly gaseous in nature and this is quite a contrast to the next few members which are solid in nature.
As we move down the group, we can see that there is an increase in the number of solids. Hence, as we go down the group we can see a relative increase in order and thus we expect more stability at room temperature compared to the volatility of the shorter chain
Explanation:
The given reaction is as follows.
Hence, number of moles of NaOH are as follows.
n = 
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n = 
= 0.0025 mol
According to ICE table,
Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] = 
[NaA] =
![[A^{-}] = [NaA] = \frac{0.0025 mol}{V}](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20%5BNaA%5D%20%3D%20%5Cfrac%7B0.0025%20mol%7D%7BV%7D)
Now, we will calculate the 
value as follows.
pH = 
![pK_{a} = pH - log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=pK_%7Ba%7D%20%3D%20pH%20-%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= 
= 3.42
Thus, we can conclude that of the weak acid is 3.42.
